Let $\mathcal{N}$ denote the set of all products of (powers of) $2,3$ and $5$:
$$ \mathcal{N} = \{ 2^a 3^b 5^c \ : \ a,b,c \geq 0 \} \subset \mathbb{N}.$$
We use the elements of $\mathcal{N}$ to partition $\mathbb{N}$ into intervals. In other words, we order $\mathcal{N} = \{N_0 < N_1 < \dots \}$ and consider the intervals $I_k = [N_k,N_{k+1})$. I will call an interval $I_k$ *exceptional* if $N_k$ is odd. Let $E$ be the union of all exceptional intervals.

Is it true that $E$ has asymptotic density zero?

For instance: $\mathcal{N} = \{1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,\dots\}$ so the few initial intervals are $I_0 = [1,2)$, $I_1 = [2,3),\ldots,I_{10} = [15,16)$, $I_{11} = [16,18)$, $I_{12} = [18,20)$, $I_{13} = [20,24)$, $I_{14} = [24,25)$, $I_{15} = [25,27)$. The intervals $[15,16)$ and $[25,27)$ are exceptional while $[16,18)$, $[18,20)$, $[20,24)$, $[24,25)$ are not exceptional. A few initial elements of $E$ are $1,3,5,9,15,25,26,27,28,29 \dots$.

Rationale: The number of points in $\mathcal{N}$ in the interval $[X,2X)$ for large $X$ is of the order of $\log^2 X$. On the other hand, the number of points in $\mathcal{N}$ that are initial points of exceptional intervals (i.e., integers of the form $3^b 5^c$) in $[X,2X)$ is of the order of $\log X$. If we knew that all of the intervals $I_k$ contained in $[X,2X)$ have roughly the same length, then clearly this would imply that $| E \cap [X,2X)| \simeq X/\log X = o(X)$, as needed. If instead of $\{2,3,5\}$ we only had $\{2,3\}$ then this type of an argument (together with the basic fact that the sequence $n \log_2 3$ is equidistributed modulo $1$) would suffice to show that $d(E) = 0$. It seems like the variant for three primes might follow from more careful analysis of expressions like $n \log_2 3 + m \log_2 5 \bmod 1$.

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